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Minimum Time For K Virus Variants to Spread

Number: 2073

Difficulty: Hard

Paid? Yes

Companies: N/A

Problem Description

Given an infinite 2D grid and n unique virus variants starting at given points, each virus spreads to its four adjacent cells every day. The variants do not interfere if they meet at the same cell. We must determine the minimum number of days required such that there exists at least one cell that is infected by at least k distinct virus variants.

Key Insights

  • Each virus spread can be seen as a Manhattan distance “ball” (diamond) with radius equal to the number of days.
  • A cell (x, y) is within the spread of a virus originating at (x_i, y_i) after d days if |x - x_i| + |y - y_i| ≤ d.
  • By transforming this Manhattan condition using new coordinates s = x + y and t = x - y, the spread of a virus becomes two 1D intervals:
    • s interval: [x_i + y_i - d, x_i + y_i + d]
    • t interval: [x_i - y_i - d, x_i - y_i + d]
  • The problem then reduces to finding a candidate point (expressed in the s-t coordinate system) that lies within at least k such intervals in both dimensions.
  • The condition is monotonic in d, making binary search an ideal strategy to find the minimum day.

Space and Time Complexity

Time Complexity: O(n^2 * log(max_distance)) – For each candidate day (via binary search) we check overlapping intervals from up to n viruses.
Space Complexity: O(n) – Used for storing intervals and candidate boundaries.

Solution

We apply binary search on the days d. For each candidate day:

  1. Compute intervals for each virus variant in transformed coordinates (s = x+y, t = x-y).
  2. Collect candidate boundary points from these intervals.
  3. For each candidate pair (one candidate from the s intervals and one from the t intervals), count how many virus intervals cover that candidate point.
  4. If any candidate point is covered by at least k viruses in both dimensions, the day d is feasible; otherwise, it is not.
  5. Adjust the binary search boundaries accordingly to locate the minimum day.

Code Solutions

def minimumDaysForKVariants(points, k):
    # Function to check if there exists a candidate point covered by at least k virus variants after d days.
    def can_find(d):
        sum_intervals = []
        diff_intervals = []
        for x, y in points:
            s = x + y
            t = x - y
            # Each virus covers an interval on the s (x+y) and t (x-y) axes
            sum_intervals.append((s - d, s + d))
            diff_intervals.append((t - d, t + d))
        
        # Collect candidate boundaries for s and t
        s_candidates = set()
        t_candidates = set()
        for L, R in sum_intervals:
            s_candidates.add(L)
            s_candidates.add(R)
        for L, R in diff_intervals:
            t_candidates.add(L)
            t_candidates.add(R)
            
        # Check if any candidate (s, t) is inside at least k intervals.
        for s_val in s_candidates:
            for t_val in t_candidates:
                count = 0
                for (sL, sR), (tL, tR) in zip(sum_intervals, diff_intervals):
                    if sL <= s_val <= sR and tL <= t_val <= tR:
                        count += 1
                    if count >= k:
                        return True
        return False

    # Binary search over d (days); hi bound is set based on worst-case Manhattan distance.
    lo, hi = 0, 250
    while lo < hi:
        mid = (lo + hi) // 2
        if can_find(mid):
            hi = mid
        else:
            lo = mid + 1
    return lo

# Example usage:
print(minimumDaysForKVariants([[1,1],[6,1]], 2))  # Expected output: 3
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